Given $\dot x(t)=3x(t)$ and $x(0)=\frac32$ what is $x(2)$? I believe this is a differential delay equation but I'm not sure. I've tried integration but that was confusing. It has also been suggested that I can find the original equation from the first derivative given but that's not right either. 
$$\frac{dx}{dt}=3x^2$$ the original would be $x^3$
 A: You have 
$$\dot x = \frac{d}{dt}x = 3x$$
multiply $dt$
$$dx = 3xdt$$
divide $x$
$$\frac{1}{x}dx = 3dt$$
integrate
$$\int \frac{1}{x}dx = \int 3dt$$
$$\ln x + C_{apples} = 3t + C_{oranges}$$
make a fruit salad
$$\ln x = 3t + C_{\text{fruit salad}}$$
exponentializify1
$$x = e^{3t + C_{\text{fruit salad}}}=e^{3t}e^{C_{\text{fruit salad}}}$$
add some chopped chocolate to the salad
$$x = e^{3t}e^{C_{\text{fruit salad}}}=e^{3t}C_{\text{fruit salad with chocolate}}$$
try the salad with $x(0)=\frac32$
$$\frac32 =e^{3\cdot 0}C_{\text{fruit salad with chocolate}}=C_{\text{fruit salad with chocolate}}$$
if you think the salad is appropriately seasoned, serve on 2
$$x(2)=e^{3\cdot 2}C_{\text{fruit salad with chocolate}} = e^6\frac32$$

1that is a word
A: Notice, we have $$x'(t)=3x(t)$$ 
$$\frac{dx}{dt}=3x$$
$$\frac{dx}{x}=3dt$$ 
$$\int \frac{dx}{x}=3\int dt$$  $$\ln |x|=3t+C$$$$\iff x=e^{3t+C}$$
Now, setting $x=\frac{3}{2}$ at $t=0$, we get 
$$\frac{3}{2}=e^{3(0)+C}\iff e^C=\frac{3}{2}\iff C=\ln\frac{3}{2}$$ 
Now, setting the value of $C$ we get 
$$x=e^{3t+\ln\frac{3}{2}}=e^{3t}e^{\ln\frac{3}{2}}=\frac{3}{2}e^{3t}$$
Now, setting $t=2$, we get $$x(2)=\frac{3}{2}e^{3(2)}=\frac{3}{2}e^{6}$$
Hence, we get $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x(2)=\frac{3}{2}e^{6}}}$$
A: $$\frac{dx(t)}{dt} = 3 x(t) \Leftrightarrow x(t) = k\cdot e^{3t}$$
Now the boundary condition:
$$ k\cdot e^0=\frac{3}{2}  \Leftrightarrow k=\frac{3}{2}$$
which yields:
$$ x(t) = \frac{3}{2}e^{3t}$$
It's now a matter of calculating this function for $t = 2$:
$$ x(2) = \frac{3}{2} e^6 \approx 605.14$$
